How do you solve the equation $a x^{2} + 5 = 3$ $ax^2+5=3$ ?

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Answer 1 · 2018-05-18T22:36:46

In terns of x:

$x = \pm \sqrt{\frac{- 2}{a}}$ $x = +-sqrt((-2)/a)$

if a is not negative the solution will be imaginary.

$a x^{2} + 5 = 3$ $ax^2 + 5 = 3$

$a x^{2} + 5 - 5 = 3 - 5$ $ax^2 + 5 -5 = 3 - 5$

$a x^{2} = - 2$ $ax^2 = -2$

$\frac{a x^{2}}{a} = \frac{- 2}{a}$ $(ax^2)/a = (-2)/a$

$x^{2} = \frac{- 2}{a}$ $x^2 = (-2)/a$

$\sqrt{x^{2}} = \pm \sqrt{- \frac{2}{a}}$ $sqrt(x^2) = +-sqrt(-2/a)$

$x = \pm \sqrt{\frac{- 2}{a}}$ $x = +-sqrt((-2)/a)$

How do you solve the equation ax^2+5=3ax2+5=3ax^2+5=3?